Two moles of a gas at a temperature of 327^{c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Number of moles $n = 2$, Initial temperature $T_1 = 327 + 273 = 600 \ K$, Ratio of specific heats $\gamma = \frac{4}{3}$.Volume increases b

Vedclass · Accepted Answer

$14.94$

Vedclass · Answer

(A) Given: Number of moles $n = 2$, Initial temperature $T_1 = 327 + 273 = 600 \ K$, Ratio of specific heats $\gamma = \frac{4}{3}$.Volume increases by $700 \%$, so $V_2 = V_1 + 7V_1 = 8V_1$.For an adiabatic process, $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.$T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma-1} = 600 \left( \frac{1}{8} ight)^{\frac{4}{3}-1} = 600 \left( \frac{1}{8} ight)^{1/3} = 600 	imes \frac{1}{2} = 300 \ K$.Work done in an adiabatic process is $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.$W = \frac{2 	imes 8.3 	imes (600 - 300)}{\frac{4}{3} - 1} = \frac{2 	imes 8.3 	imes 300}{1/3} = 2 	imes 8.3 	imes 300 	imes 3 = 14940 \ J = 14.94 \ kJ$.

Two moles of a gas at a temperature of $327^{\circ} C$ expands adiabatically such that its volume increases by $700 \%$. If the ratio of the specific heat capacities of the gas is $\frac{4}{3}$, then the work done by the gas is (Universal gas constant $= 8.3 \ J \ mol^{-1} \ K^{-1}$) (in $kJ$)

Similar Questions

The work of $146 \ kJ$ is performed in order to compress one kilo mole of gas adiabatically,and in this process,the temperature of the gas increases by $7 ^\circ C$. The gas is $(R = 8.3 \ J \ mol^{-1} K^{-1})$.

The relation obeyed by a perfect gas during an adiabatic process is $PV^{3/2} = \text{constant}$. The initial temperature of the gas is $T$. When the gas is compressed to half of its initial volume, the final temperature of the gas is:

The work done in which of the following processes is equal to the internal energy of the system?

In an adiabatic expansion of a gas, the initial and final temperatures are $T_1$ and $T_2$ respectively. Then, the change in internal energy of the gas is:
$[R = \text{gas constant}, \gamma = \text{adiabatic ratio}]$

During an adiabatic process,the pressure of a gas is found to be proportional to the cube of its temperature. The ratio $C_P / C_V$ for this gas is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module